Can someone describe a simple experiment to convince a person thinking about physics classically (called Claus) that quantum mechanics has something weird, entangled?

I mean an experiment that he wouldn't been able to explain classically. It should not be one of these proofs that introduces abstract equations and frameworks, derives inequalities and argues indirectly. It should be down to earth such as saying:

"Here is the experiment... Here is a possible outcome of measurements after N experiments...AABBAAB... Please propose a classical (hidden variable) explanation to reproduce this or convince yourself of the impossibility otherwise."

Basically the goal is to explain the effect in his words and thinking. For a particular setting Claus will try to propose a classical explanation. He would like to be proven wrong rather than asked to examine an indirect argument of the quantum mechanics.

I know basic quantum mechanics and can fill the calculation gaps. I've seen common Bell inequalities but find it hard to extract such a classical-view-only argument.

  • 2
    $\begingroup$ If you haven't already, take a look at Sid Coleman's Quantum Mechanics In Your Face lecture. It describes a very explicit experiment and shows results which would be obtained. You could try to see if you could somehow explain those results classically and see where this leads... $\endgroup$
    – twistor59
    Jun 4, 2012 at 11:37
  • $\begingroup$ Thanks, talks exactly about this question. The answer of Frédéric describes it a bit more, too. $\endgroup$
    – Gere
    Jun 5, 2012 at 7:00
  • $\begingroup$ Good exposition of Mermin's paradigm. $\endgroup$ Jan 28, 2021 at 19:09

2 Answers 2


What you are looking for is an experiment violating a Bell inequality, like the CHSH inequality. From the redaction of your question, I infer that you've already looked at it but weren't convinced (Tell me if I'm wrong.) Maybe the Memin-GHZ game will convince you.


It is a game where 3 parties (A,B,C) play together against a referee (R). The nice aspect of this game is that A,B and C can win this game with probability 100%, while the best classical strategy has a probability of winning of 75%. The "down to earth" description of the experiment could then be reduced to :

  • Play the game $N$ times, with $N\gg4$
  • The results are A,B,C win every time (or at least with a probability significantly greater than 75 %)
  • The only possible explanations for Claus are then :
    • A, B, C do not obey classical laws
    • A, B, and C cheated, either by communicating during the game or by bribing the referee. A suitable design of the experiment could bake these probabilities unlikely.

This experiment has been performed, but I don't know the reference by heart. (Edit: The paper is : Jian-Wei Pan, Dik Bouwmeester, Matthew Daniell, Harald Weinfurter & Anton Zeilinger, Experimental test of quantum nonlocality in three-photon Greenberger-Horne-Zeilinger entanglement Nature 403 (2000) 515-519, available here.)

Rules of the Mermin-GHZ game

The rule of the game are the following :

  • The game is played in many rounds;
  • A, B, C can only communicate before a round starts. Any communication between them during a round is prohibited;
  • During a round, R asks individually a question to each of A,B,C. This question is taken among the two following : $X$,$Y$
  • To this question, A, B, C's answer can be either $+1$ or $-1$
  • R is not allowed to ask any combinations of $X$ and $Y$s. The set of allowed questions are $XYY$, $YXY$, $YYX$ and $XXX$.
  • If the questions are $XXX$, A, B, and C gain if the product of their answers is $-1$. Else, they win if this product is $+1$.

Impossibility of a winning classical strategy

A classical deterministic strategy can described by the answer given by each participant to each possible question, i.e. by 6 numbers $X_A, Y_A, X_B, Y_B, X_C, Y_C$. For the strategy to be always winning, one need to obey the following equalities : $$\begin{align} X_A X_B X_C &=-1 \\ X_A Y_B Y_C &=+1 \\ Y_A X_B Y_C &=+1 \\ Y_A Y_B X_C &=+1 \end{align}$$
Some algebra can easily show that at most three of this equalities can be satisfied at the same time.

A classical probabilistic strategy can be described as the random choice between deterministic strategy and can therefore not be better than a deterministic strategy.

Is there a quantum strategy ?

Yes, quantum A, B, and C can always win the game. I won't detail the strategy here, because it wouldn't interest Claus ;-)

  • $\begingroup$ Thanks! This looks quite promising and I have to study this closer! I read the CHSH inequality, but havent got a deeper grasp on it. When a layman asked me what I'm trying to understand, I knew which way I wanted to answer, but wasn't sure how to frame the CHSH this way. $\endgroup$
    – Gere
    Jun 4, 2012 at 14:37
  • 2
    $\begingroup$ To phrase the CHSH inequality this way, the usual formulation is the CHSH game : Alice and Bob receive binary (0,1) questions ($x$,$y$). their binary answer are $a$ and $b$. They win if $a\oplus b=xy$. The best quantum strategy wins 85% of the time, while the best classical only 75% $\endgroup$ Jun 4, 2012 at 14:45
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    $\begingroup$ This is interesting, I think I have read something similar in Greene's book The Fabric of the Cosmos. Could you please clear up the last sentence? It does not make sense to me. $\endgroup$ Jun 4, 2012 at 14:47
  • $\begingroup$ @IstvánZachar : Sorry for the typos. It should be corrected now. $\endgroup$ Jun 4, 2012 at 14:51
  • 1
    $\begingroup$ Thanks Frédéric, I did find the very same experiment of Bell in The Fabric of the Cosmos, described as a very clear-cut and understandable explanation for the layman, at page 105. $\endgroup$ Jun 4, 2012 at 16:38

The easiest example is the small-angle limit of the classical Bell inequality, which is probably the case where Bell noticed the violation first, since he mentions it explicitly as a particularly intuitive limit in his paper.

The situation is this: I have a machine that sends out two particles to two far away detectors. I can measure a yes/no quantity in 1 of 3 directions A,B,C. I notice the following:

  • A,B,C always give the same answer on the two particles, so that they are 100% correlated.
  • If I measure A on one particle and B on the other, the results are 99% correlated, so that 1% of the answers are different.
  • If I measure B on one particle and C on the other, the results are 99% correlated, so taht only 1% of the answers are different between B and C.

Thinking classically, there are some variables determining the outcome of the measurements. Whatever these variables, they are the same in the two particles regarding the outcome of A,B,C (so as to reproduce point 1), and they are 99% correlated between the outcome of A and B, and between the outcomes of B and C.

This means that 99% of the time, the variable for A is the same as the variable for B, and 99% of the time the variable for B is the same as the variable for C. The largest fraction of the time that C can be different from A is then 98%, assuming that all the mismatching situations are disjoint.

But in quantum mechanics, when you do this, the results are 96% correlated! This might look like a small violation, but it is double the maximum allowed mismatch, so it's proportionately a big violation.

The experimental way to set this up is to consider a spin state of two electrons where they are fully entangled in a spin-singlet, so that the measurement of spin in three nearby directions A,B,C are always opposite. In order to make the situation exactly parallel to what I said above, you have to flip the results on one of the particles to make them 100% correlated instead of anti-correlated.

Now measuring one particle collapses the other to a definite spin state, and the probability of getting a mismatched spin in direction $\theta$ goes as the |sin(\theta/2)|^2 which goes as $|\theta^2|$. if you adjust the $\theta$ to make the probability of A and B mismatching 1%, you have the same $\theta$ between B and C, and so the angle between A and C is doubled. When you double $\theta$, you get 4 times the mismatch, because the probability goes as the angle squared.

From this you see that the problem is fundamental to the squaring of amplitudes, and any fix would require a mismatch in probability which is a sharp-turn type thing like $|\theta|$, where doubling the angle doubles the mismatch. This is clearly incompatible with the basis rotational invariance in quantum mechanics.

This is the intuitively simplest violation, in the spirit of Bell. Mermin presented things much in this way, but the small angle limit is important, because it makes the mathematics obvious as day.


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